-
Algebraic Structure
Let x = ( x, ... , x ), y = ( y, ... , y ) ∈ Rn and α ∈ R.
1) sum x + y := ( x1 + y1, x2+ y2, ... , xn + yn )
2) difference x - y := ( x1 - y1, x2 - y2, ... , xn - yn )
3) product αx := ( αx1, αx2, ... , αx3 )
4) dot product x · y : = x1y1 + x2y2 + ... + xnyn
-
Algebraic Definition 1
i) Euclidean Norm
ii) L-one-norm
iii) sup-norm
iv) distance
-
Algebraic Definition
i) Euclidean Norm : 
ii)
iii)
-
Algebraic Definition:
For a, b,
i) Orthogonality
ii) Parallel
i) a and b are said to be parallel if and only if there is a scalar t ∈ R s.t. a = tb
ii) a and b are said to be orthogonal if and only if a b = 0.
-
Algebraic Structure : Inequality
-
-
Topology of Rn: Definition
i) open ball
ii) closed ball
i) For each r > 0, the open ball centered at a of radius r is the set of points
ii) For each , the closed ballat centered at a of radius r is the set of points
-
Topology of Rn: Definition
i) Open set
ii) Closed set
i) A subset V of R n is said to be open (in R n)
- ii) A subset V of Rn is said to be closed (in Rn)
-
-
Topology: Remark (8.21)
Prove:
Let . Set . If , then by the Triangle Inequality and the choice of e,
Thus, by definition, . In particular,
-
Topology: Remark (8.22)
Prove:
Let and set . Then, by definition, , so . Therefore, E c is open.
-
Topology: Remark (8.22)
Prove:
For each n ∈ N, the empty set ∅ and the whole space Rn are both open and closed.
Since Rn = ∅c and ∅ = (Rn)c, suffices to prove that ∅ and Rn are both open.
Since the empty set contains no points, "every" point x ∈ ∅ satisfies Be(x) ⊆ ∅ (vacuously). Therefore, ∅ is open.
On the other hand, since Be(x) ⊆ Rn, ∀ x ∈ Rn and e > 0, Rn is open
-
Topology : Theorem
If
i) {Vα}α∈A any collection of open subsets of Rn
ii) {Vk : k=1, 2, ..., p} a finite collection of open subsets of Rn
iii) {Eα}α∈Aany collection of closed subsets of Rn
iv) {Ek : k = 1, 2, ... , p} a finite collection of closed subsets of Rn,
v) V is open and E is closed,
then
i) is open
ii) is open iii)
-
|
|
